How to solve #a^3-3a+2=0# ?

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Answer 1 · 2018-03-31T13:26:18

(a-1)(a-1)(a+2)

#a^3 -3a+2#
#a^3 -a^2+a^2-a-2a+2#
#a^2(a-1)+a(a-1)-2(a-1)#
#(a-1)(a^2+a-2)#
#(a-1)(a^2+2a-a-2)#
#(a-1)(a(a-1)-1(a+2))#
#(a-1)(a-1)(a+2)#

i dont know how to explain properly but here is the answer

Answer 2 · 2018-03-31T13:26:48

#a=1,-2#

#a^3-3a+2=0#

Any of the factors of #2# could be a value of #a#. We start with #a=1#

#1^3-3+2=0#

S #a-1# is a factor.

#a^3-3a+2=(a-1)(a^2+2a-2)=(a-1)(a-1)(a+2)=(a-1)^2(a+2)=0#

So #a=1# or #a=-2#